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If imaginary numbers bisect real numbers at zero, am I correct in thinking $0i = 0$?

And if so, does that mean $i = 0/0$?

EDIT:

I’ve just realised my careless error in my thought process.

I immediately just “removed” the $0$ from both sides without thinking about the fact that I was actually simplifying $\frac{0i}{0}=\frac{0}{0}$ to $i=\frac{0}{0}$ despite $\frac{0}{0}\neq1$, meaning $\frac{0}{0}i \neq i$

100pxsquared
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3 Answers3

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An interesting way of looking at this question is to ask the following (to expand on Lulu's answer: If we have $0\cdot 1 = 0$, why is $1\neq \frac{0}{0}$? The answer is, of course, that $\mathbb{R}$ is a field, and the zero element has no inverse. Therefore, from a pure algebraic perspective, $\frac00$ makes no sense.

Now we can ask ourselves the same question in $\mathbb{C}$. Is $\mathbb{C}$ a field? What does that imply for your question?

bliipbluup
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  • Thank you for your detailed answer, but as I commented on Parcly's answer, "According to wikipedia i^2 = -1, does that mean i = sqrt(-1)? Which as far as i know isnt possible?" – 100pxsquared Feb 03 '20 at 15:01
  • Why do you think it's not possible? May I ask what mathematical background you have? – bliipbluup Feb 03 '20 at 15:03
  • Just started A level pure maths 1 recently. – 100pxsquared Feb 03 '20 at 15:04
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    Ok. In that case, just think of $i$ as an abstract symbol defined exactly so serve the purpose of solving the equation $i^2=-1$. That is, we're expanding the numbers so that it is allowed. – bliipbluup Feb 03 '20 at 15:06
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    As you will later see, if you pursue mathematics further, this definition of $i$ has a more formal technique behind it. Namely that $\mathbb{C}\cong \mathbb{R}[X]/(X^2+1)$. We say that the complex numbers are isomorphic to the latter construction. – bliipbluup Feb 03 '20 at 15:07
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    @UnexpectedExpectation That's a bad idea, since we then get to the question of $1 = \sqrt{1} = \sqrt{-1\cdot-1}=\sqrt{-1}\cdot\sqrt{-1} = i\cdot i = -1$. It's better to think of $i$ as the symbol whose purpose is that $i^2=-1$. – 5xum Feb 03 '20 at 15:07
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    @5xum You are very right, sir. I will correct my comment. – bliipbluup Feb 03 '20 at 15:08
  • @UnexpectedExpectation Thank you for explaining it fully! (at least as much as I can understand) – 100pxsquared Feb 03 '20 at 15:09
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Yes, $0i=0$, but the prohibition on dividing by zero remains valid in the complex plane, and one does not conclude $i=0/0$.

Parcly Taxel
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It's also true that $0\cdot 1 = 0$. Does that mean $1=\frac00$?

It's also true that $0\cdot 2 = 0$. Does that mean $2=\frac00$?

5xum
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